The pressure of a gas kept in an isothermal container is 200Kpa. If half the gas is removed from it, the pressure will be

Options

A. (a)`100kPa`
B. (b)`200kPa`
C. (c)`400kPa`
D. (d)`800kPa`

Answer

To solve the problem, we will use the ideal gas law and the concept of isothermal processes. Here’s the step-by-step solution:

### Step 1: Understand the Ideal Gas Law
The ideal gas law is given by the equation:
\[ PV = nRT \]
Where:
- \( P \) = Pressure
- \( V \) = Volume
- \( n \) = Number of moles of gas
- \( R \) = Universal gas constant
- \( T \) = Temperature

### Step 2: Identify the Conditions
In this problem:
- The gas is in an isothermal container, meaning the temperature \( T \) is constant.
- The volume \( V \) of the container is also constant since it is fixed.

### Step 3: Relate Pressure and Number of Moles
From the ideal gas law, we can rearrange the equation to express pressure in terms of the number of moles:
\[ P = \frac{nRT}{V} \]
Since \( R \), \( T \), and \( V \) are constants, we can see that pressure \( P \) is directly proportional to the number of moles \( n \):
\[ P \propto n \]

### Step 4: Analyze the Change in Moles
Initially, let the number of moles of gas be \( n_1 \) and the initial pressure be \( P_1 = 200 \, \text{kPa} \). When half of the gas is removed, the new number of moles \( n_2 \) will be:
\[ n_2 = \frac{1}{2} n_1 \]

### Step 5: Calculate the New Pressure
Using the direct proportionality:
\[ \frac{P_2}{P_1} = \frac{n_2}{n_1} \]
Substituting \( n_2 \):
\[ \frac{P_2}{200 \, \text{kPa}} = \frac{\frac{1}{2} n_1}{n_1} \]
This simplifies to:
\[ P_2 = \frac{1}{2} \times 200 \, \text{kPa} \]
\[ P_2 = 100 \, \text{kPa} \]

### Step 6: Conclusion
Thus, the pressure after removing half of the gas will be:
\[ P_2 = 100 \, \text{kPa} \]

### Final Answer
The final pressure will be **100 kPa**.

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